IPMAT Indore 2020 — MCQ

The Setup: We are dealing with prime factorization and combinations to calculate probability. We need to find the total number of factors of $10^{19}$ (the sample space) and the total number of factors that carry at least $10^{15}$ in their DNA (the winning outcomes).
Step 1: Break down the base stats. Prime factorize the main number:

$$10^{19}=(2\times 5)^{19}=2^{19}\times 5^{19}$$

The formula for the total number of factors of $p^a\times q^b$ is $(a+1)(b+1)$, since each exponent is chosen freely:

$$\text{Total factors}=(19+1)(19+1)=20\times 20=400$$

Step 2: Isolate the target condition. A factor of $10^{19}$ has the shape $2^x\times 5^y$ with $0\leq x,y\leq 19$. For it to be a multiple of $10^{15}=2^{15}\times 5^{15}$, it must carry at least fifteen of each prime, so $x\geq 15$ and $y\geq 15$.
Combined with the ceiling of 19, the valid range for both $x$ and $y$ is $\{15,16,17,18,19\}$.
Step 3: Count the winning outcomes. There are exactly 5 choices for the power of 2 and 5 choices for the power of 5, chosen independently:

$$\text{Valid factors}=5\times 5=25$$

Step 4: Calculate the final RNG. Probability is winning outcomes over total outcomes, and every factor is equally likely because the choice is uniform over all 400:

$$P=\frac{25}{400}=\frac{1}{16}$$

Final Answer: $\frac{1}{16}$

The Setup: This geometry problem is a triple filter. The sides must pass the triangle inequality vibe check, then survive the acute-angle condition, then fit under the perimeter cap. Whichever filter is strictest wins.
Step 1: Set up the side lengths. Let the three consecutive integer sides be $x-1$, $x$ and $x+1$, with $x-1\geq 1$.
For any triangle the two smaller sides must beat the largest: $(x-1)+x>x+1\implies x>2$.
Step 2: Enforce the acute angle rule. A triangle is acute when *every* angle is acute - but we only have to test one. The largest angle always sits opposite the longest side, so if that angle is acute the two smaller ones are smaller still and acute automatically. The longest side here is $x+1$, so the single condition to check is $a^2+b^2>c^2$:

$$(x-1)^2+x^2>(x+1)^2$$

$$x^2-2x+1+x^2>x^2+2x+1$$

$$x^2>4x$$

Since $x$ is positive we may divide by $x$, giving $x>4$. As $x$ is an integer, the lowest valid value is $x=5$ (the smallest qualifying triangle being $4,5,6$). Note this condition is strictly tighter than Step 1's $x>2$, so the triangle inequality is absorbed and adds nothing further.
Step 3: Apply the perimeter cap. The perimeter is at most 100:

$$(x-1)+x+(x+1)=3x\leq 100 \implies x\leq 33.33\ldots$$

Since $x$ is an integer, the maximum is $x=33$, giving sides $32,33,34$ and a perimeter of $99$.
Step 4: Count the surviving squad. The valid range for $x$ runs from 5 to 33 inclusive:

$$33-5+1=29$$

Final Answer: 29

The Setup: This is a coordinate geometry speedrun using intercept form. The line crosses the x-axis at some point $A(a,0)$ and the y-axis at $B(0,b)$. Saying $M$ divides the portion between the axes into two equal halves is just a fancy way of saying $M$ is the midpoint of segment $AB$.
Step 1: Apply the midpoint formula. The midpoint of $A(a,0)$ and $B(0,b)$ is $\left(\frac{a}{2},\frac{b}{2}\right)$, and we are told that point is $M(-5,4)$:

$$\frac{a}{2}=-5\implies a=-10$$

$$\frac{b}{2}=4\implies b=8$$

Step 2: Construct the line equation. Use the two-intercept form $\frac{x}{a}+\frac{y}{b}=1$:

$$\frac{x}{-10}+\frac{y}{8}=1$$

Step 3: Clear the fractions in one shot. Multiply every term by $80$ (a common multiple of 10 and 8 chosen to land directly on the answer's scale):

$$80\cdot\frac{x}{-10}+80\cdot\frac{y}{8}=80 \implies -8x+10y=80$$

Reordering gives $10y-8x=80$.
Step 4: Verify against all three known points. A midpoint slip is easy to make, so test the line on the two intercepts and $M$ itself:
* $A(-10,0)$: $10(0)-8(-10)=80$ ✓
* $B(0,8)$: $10(8)-8(0)=80$ ✓
* $M(-5,4)$: $10(4)-8(-5)=40+40=80$ ✓

All three sit on the line, and $M$ is genuinely halfway between $A$ and $B$. Running the other three options through $M(-5,4)$ fails every time, so the match is unique.

Final Answer: $10y - 8x = 80$

The Setup: This trig expression looks intimidating, but it has zero plot armour once you spot the symmetry in the angles. We use the supplementary angle identity to fold the expression in half, then the complementary identity to finish it off.
Step 1: Fold the back half. Notice that $\frac{7\pi}{8}=\pi-\frac{\pi}{8}$, and $\cos(\pi-\theta)=-\cos\theta$, so $\cos\left(\frac{7\pi}{8}\right)=-\cos\left(\frac{\pi}{8}\right)$. Squaring kills the minus sign:

$$\cos^2\left(\frac{7\pi}{8}\right)=\cos^2\left(\frac{\pi}{8}\right)$$

Identically, $\frac{5\pi}{8}=\pi-\frac{3\pi}{8}$, so $\cos^2\left(\frac{5\pi}{8}\right)=\cos^2\left(\frac{3\pi}{8}\right)$.
Step 2: Simplify the sum. The four terms collapse into two identical pairs:

$$2\left[\cos^2\left(\frac{\pi}{8}\right)+\cos^2\left(\frac{3\pi}{8}\right)\right]$$

Step 3: Exploit the complementary angles. Notice that $\frac{\pi}{8}+\frac{3\pi}{8}=\frac{4\pi}{8}=\frac{\pi}{2}$. The two angles are complementary, and $\cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta$, so:

$$\cos\left(\frac{3\pi}{8}\right)=\sin\left(\frac{\pi}{8}\right)$$

Step 4: Use the legendary identity. Substitute sine into the folded expression:

$$2\left[\cos^2\left(\frac{\pi}{8}\right)+\sin^2\left(\frac{\pi}{8}\right)\right]$$

By the Pythagorean identity $\cos^2\theta+\sin^2\theta=1$, the bracket collapses to 1:

$$2\times 1=2$$

Final Answer: $2$

The Setup: We are manipulating the famous Basel problem. We are handed the sum of *all* inverse squares and asked for the sum of only the *odd* ones. The trick is to split the total into its even and odd halves, then notice the even half is a scaled copy of the whole thing.
Step 1: Define the series. Let the total be $S=\frac{\pi^2}{6}$. Every positive integer is either odd or even, so the total splits cleanly with nothing counted twice and nothing missed:

$$S=\text{Sum(Odds)}+\text{Sum(Evens)}$$

Every term here is positive and the series converges, so splitting and rearranging it like this is legitimate rather than a formal trick.
Step 2: Factor the even series. Write out the even terms and pull out the common factor. Each even number is $2n$, so its square is $4n^2$:

$$\text{Sum(Evens)}=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\ldots=\frac{1}{4}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots\right)$$

The bracket is literally the original total $S$, so $\text{Sum(Evens)}=\frac{S}{4}$. This self-similarity is the whole engine of the problem.
Step 3: Solve for the odds. Substitute back into the master equation:

$$S=\text{Sum(Odds)}+\frac{S}{4} \implies \text{Sum(Odds)}=S-\frac{S}{4}=\frac{3S}{4}$$

Step 4: Final substitution. Plug in the known value of $S$:

$$\text{Sum(Odds)}=\frac{3}{4}\times \frac{\pi^2}{6}=\frac{3\pi^2}{24}=\frac{\pi^2}{8}$$

Final Answer: $\frac{\pi^2}{8}$

The Setup: This is a classic Bayes' Theorem trap. We have a prior probability (the physical die) mixed with a conditional one (the man's honesty stats). We want the probability that a five *actually* came up, given that he claims it did - which is not the same as his truthfulness rate, and the gap between $\frac{4}{5}$ and the answer is the entire point of the question.
Step 1: Map the base probabilities.
* Probability of rolling a five: $P(A)=\frac{1}{6}$
* Probability of rolling anything else: $P(\text{not }A)=\frac{5}{6}$
* Man tells the truth: $P(T)=\frac{4}{5}$
* Man lies: $P(L)=\frac{1}{5}$

Step 2: Map the conditional claims.
If it genuinely is a five, he reports five precisely when he is being truthful: $P(\text{Claims 5}\mid \text{Is 5})=\frac{4}{5}$.
If it is not a five, he can only report five by lying, so this branch is taken as $P(\text{Claims 5}\mid \text{Not 5})=\frac{1}{5}$ - the exam's convention is that a lie about the target face lands on that face.
*Worth naming the assumption, because it is doing real work.* If instead a liar spread his lie uniformly across the five wrong faces, this branch would be $\frac{1}{5}\times\frac{1}{5}=\frac{1}{25}$ and the answer would come out as $\frac{4}{5}$. That value appears nowhere in the options, whereas the convention above lands exactly on option 1 - so the option set itself tells us which reading the paper intended.
Step 3: Apply Bayes' Theorem.

$$P(\text{Is 5}\mid \text{Claims 5})=\frac{P(\text{Is 5})\times P(\text{Claims 5}\mid \text{Is 5})}{P(\text{Is 5})\times P(\text{Claims 5}\mid \text{Is 5})+P(\text{Not 5})\times P(\text{Claims 5}\mid \text{Not 5})}$$

Step 4: Crunch the numbers.

$$\text{Numerator}=\frac{1}{6}\times \frac{4}{5}=\frac{4}{30}$$

$$\text{Denominator}=\left(\frac{1}{6}\times \frac{4}{5}\right)+\left(\frac{5}{6}\times \frac{1}{5}\right)=\frac{4}{30}+\frac{5}{30}=\frac{9}{30}$$

$$P=\frac{\ \frac{4}{30}\ }{\frac{9}{30}}=\frac{4}{9}$$

Sanity check on the direction: $\frac{4}{9}<\frac{1}{2}$, which is right - a five is rare enough on a fair die that even a mostly-honest report leaves it more likely than not that he is lying.

Final Answer: $\frac{4}{9}$

The Setup: This is a logarithm inception. We unpack it from the outside in using the fundamental definition $\log_b(A)=C\implies A=b^C$, then check the candidates against the option list.
Step 1: Destroy the outer log.

$$\log_5\left(\log_8(x^2-1)\right)=0$$

Convert from log form to exponential form:

$$\log_8(x^2-1)=5^0=1$$

Step 2: Destroy the inner log. Apply the same rule to what remains:

$$x^2-1=8^1=8$$

Step 3: Solve the quadratic.

$$x^2=9 \implies x=\pm 3$$

Step 4: Pick the value that is actually on the menu. Both $x=3$ and $x=-3$ satisfy the equation - the expression only ever sees $x^2$, so the sign is irrelevant, and $x^2-1=8>0$ keeps every logarithm well defined. This is exactly why the stem asks for *a possible* value rather than *the* value. Of the two, only $3$ appears among the options, so it is the answer.
The other three options all fail, and one of them is a genuine trap:
* $x=2\sqrt{2}$: $x^2-1=7$, and $\log_8 7\approx 0.936$, so the outer log is not 0.
* $x=\sqrt{2}$: $x^2-1=1$, so $\log_8 1=0$ - and $\log_5 0$ is undefined, not zero. Mistaking one for the other is the intended snare.
* $x=2$: $x^2-1=3$, and $\log_8 3\approx 0.528$, so the outer log is not 0.

Final Answer: 3

The Setup: Cross-multiplying an inequality whose denominator changes sign is how this question eats people alive. We dodge that entirely with the sum of cubes identity $a^3+b^3=(a+b)(a^2-ab+b^2)$, taking $a=1$ and $b=x$, which turns the whole comparison into a single fraction whose sign we can read off.
Step 0: Note the structure. Statements (i) and (ii) are exact opposites on the same interval, as are (iii) and (iv). So exactly one of each pair must be true, and the answer is guaranteed to be one statement from each pair - that alone kills option 1 ((i) and (ii) cannot both hold) before any algebra.
Step 1: Exploit the identity. With $a=1$, $b=x$:

$$1+x^3=(1+x)(1-x+x^2)$$

On both intervals $x>-1$, so $1+x>0$ - it is never zero, and we may safely divide by it:

$$1-x+x^2=\frac{1+x^3}{1+x}$$

Step 2: Find the delta. Subtract the two quantities being compared and put them over the common denominator:

$$(1-x+x^2)-\frac{1}{1+x}=\frac{1+x^3}{1+x}-\frac{1}{1+x}=\frac{x^3}{1+x}$$

Everything now rides on the sign of $\frac{x^3}{1+x}$, and since $1+x>0$ throughout, that sign is simply the sign of $x^3$ - which is the sign of $x$.
**Step 3: Evaluate Case 1, $0<x<1$.** The numerator $x^3$ is positive and the denominator $1+x$ is positive, so the delta is positive:

$$\frac{x^3}{1+x}>0 \implies 1-x+x^2>\frac{1}{1+x}$$

Flipping that around to the form the paper uses gives $\frac{1}{1+x}<1-x+x^2$, which is exactly Statement (i) - and therefore (ii) is false.
*Spot check at $x=\frac{1}{2}$:* $\frac{1}{1+x}=\frac{2}{3}$ and $1-x+x^2=\frac{3}{4}$, and indeed $\frac{2}{3}<\frac{3}{4}$, with the gap $\frac{1}{12}$ matching $\frac{x^3}{1+x}=\frac{1/8}{3/2}=\frac{1}{12}$.
**Step 4: Evaluate Case 2, $-1<x<0$.** Here the numerator $x^3$ is negative (a negative number cubed stays negative), while the denominator $1+x$ is still positive - for instance $1+(-0.5)=0.5$. A negative over a positive is negative:

$$\frac{x^3}{1+x}<0 \implies 1-x+x^2<\frac{1}{1+x}$$

Flipping it round gives $\frac{1}{1+x}>1-x+x^2$, which is exactly Statement (iv) - and therefore (iii) is false.
*Spot check at $x=-\frac{1}{2}$:* $\frac{1}{1+x}=2$ and $1-x+x^2=\frac{7}{4}$, and indeed $2>\frac{7}{4}$, with the gap $-\frac{1}{4}$ matching $\frac{x^3}{1+x}=\frac{-1/8}{1/2}=-\frac{1}{4}$.
Step 5: Assemble. The true statements are (i) and (iv), one from each contradictory pair as Step 0 predicted.

Final Answer: (i) and (iv)

The Setup: Mixture problems just require tracking the absolute volumes of the components like inventory. We find the exact litres of water and milk we currently hold, merge them, then solve for the extra water needed to hit the target 25%. The key discipline is that adding pure water changes the water column *and* the total, but leaves the milk column frozen.
Step 1: Audit Mixture 1. Total $=50$ L, water is 30%.

$$\text{Water}=0.30\times 50=15\text{ L}, \qquad \text{Milk}=50-15=35\text{ L}$$

Step 2: Audit Mixture 2. Total $=80$ L, water is 20%.

$$\text{Water}=0.20\times 80=16\text{ L}, \qquad \text{Milk}=80-16=64\text{ L}$$

Step 3: Merge the inventory. Tip both into the same barrel:

$$\text{Total volume}=50+80=130\text{ L}, \quad \text{Water}=15+16=31\text{ L}, \quad \text{Milk}=35+64=99\text{ L}$$

Step 4: Calculate the required additive. Let $w$ litres of pure water go in. Both the water and the total grow by $w$, and a 25% water mix means water is $\frac{1}{4}$ of the total:

$$\frac{31+w}{130+w}=\frac{1}{4}$$

Cross-multiply and solve:

$$4(31+w)=130+w \implies 124+4w=130+w \implies 3w=6 \implies w=2$$

Step 5: Verify from the milk side. The milk column never moved, so it is an independent check. Final total $=130+2=132$ L, of which milk is still 99 L:

$$\frac{99}{132}=\frac{3}{4}=75\% \text{ milk}, \text{ hence } 25\% \text{ water}$$

The two routes agree, so $w=2$ is right.

Final Answer: 2

The Setup: This is a rates problem, and the golden rule is that *times do not add, rates do*. We convert *hours to build the wall* into *walls per hour*, sum the three rates to 1 wall/hour, and only convert back to time at the very end.
Step 1: Define the rates. Let $r_1,r_2,r_3$ be the workers' rates in walls per hour, and $t_i=\frac{1}{r_i}$ their solo times.
All three together take 1 hour, so their rates sum to one whole wall per hour:

$$r_1+r_2+r_3=\frac{1}{1}=1$$

Worker 1 is twice as fast as Worker 3, so $r_1=2r_3$, i.e. $r_3=\frac{r_1}{2}$ (equivalently $t_3=2t_1$).
Step 2: Link Workers 1 and 2. Worker 1 finishes an hour sooner than Worker 2, and that condition is about *times*, so it must be written in times first:

$$t_1=t_2-1 \implies \frac{1}{r_1}=\frac{1}{r_2}-1 \implies \frac{1}{r_2}=\frac{1}{r_1}+1=\frac{1+r_1}{r_1}$$

Flipping gives $r_2=\frac{r_1}{1+r_1}$.
Step 3: Solve the master rate equation. Substitute $r_2$ and $r_3$ into the sum:

$$r_1+\frac{r_1}{1+r_1}+\frac{r_1}{2}=1$$

Multiply through by $2(1+r_1)$ to clear denominators:

$$2r_1(1+r_1)+2r_1+r_1(1+r_1)=2(1+r_1)$$

$$2r_1+2r_1^2+2r_1+r_1+r_1^2=2+2r_1$$

$$3r_1^2+5r_1=2+2r_1 \implies 3r_1^2+3r_1-2=0$$

By the quadratic formula:

$$r_1=\frac{-3\pm\sqrt{9-4(3)(-2)}}{6}=\frac{-3\pm\sqrt{33}}{6}$$

A rate cannot be negative, and $\sqrt{33}\approx 5.745$, so the minus root is discarded and $r_1=\frac{\sqrt{33}-3}{6}$.
Step 4: Convert back to times. Rationalising the denominator each time:

$$t_1=\frac{1}{r_1}=\frac{6}{\sqrt{33}-3}=\frac{6(\sqrt{33}+3)}{33-9}=\frac{6(\sqrt{33}+3)}{24}=\frac{\sqrt{33}+3}{4}$$

$$t_2=t_1+1=\frac{\sqrt{33}+3}{4}+\frac{4}{4}=\frac{\sqrt{33}+7}{4}, \qquad t_3=2t_1=\frac{2\sqrt{33}+6}{4}$$

Step 5: Average them. Note the question asks for the average of the *times*, not of the rates - averaging rates instead is the single most common way to lose this mark.

$$t_1+t_2+t_3=\frac{(\sqrt{33}+3)+(\sqrt{33}+7)+(2\sqrt{33}+6)}{4}=\frac{4\sqrt{33}+16}{4}=\sqrt{33}+4$$

$$\text{Average}=\frac{\sqrt{33}+4}{3}$$

Step 6: Sanity check the numbers. Numerically $t_1\approx 2.19$, $t_2\approx 3.19$, $t_3\approx 4.37$ hours. Every one exceeds 1 hour, as it must - a lone worker cannot beat all three together. Their gap $t_2-t_1$ is exactly 1, $t_3$ is exactly double $t_1$, and $\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}=1$ to twelve decimal places. All three given conditions hold.

Final Answer: $\dfrac{\sqrt{33} + 4}{3}$

The Setup: This is Inclusion-Exclusion. We count the students playing at least one sport (the union of the three sets), then subtract from 140 to get the NPCs playing nothing. Note the sets overlap heavily - a roll number like 30 is in all three - which is exactly why we cannot simply add.
Step 1: Calculate the single sets. Counting multiples of $k$ up to 140 is $\left\lfloor\frac{140}{k}\right\rfloor$.

$$|A|\ (\text{cricket, }2)=\left\lfloor\frac{140}{2}\right\rfloor=70, \quad |B|\ (\text{football, }5)=\left\lfloor\frac{140}{5}\right\rfloor=28, \quad |C|\ (\text{basketball, }3)=\left\lfloor\frac{140}{3}\right\rfloor=46$$

Step 2: Calculate the pairwise intersections. A student in two sets has a roll number divisible by both divisors, so the correct counter is their LCM - not their product. Here $2,3,5$ are pairwise coprime, so the LCM happens to equal the product, but that coincidence is what makes this step safe; with divisors like 2 and 4 the product would give a wrong count.

$$|A\cap B|\ (\text{LCM}=10)=\left\lfloor\frac{140}{10}\right\rfloor=14, \quad |A\cap C|\ (\text{LCM}=6)=\left\lfloor\frac{140}{6}\right\rfloor=23, \quad |B\cap C|\ (\text{LCM}=15)=\left\lfloor\frac{140}{15}\right\rfloor=9$$

Step 3: Calculate the triple intersection.

$$|A\cap B\cap C|\ (\text{LCM}=30)=\left\lfloor\frac{140}{30}\right\rfloor=4$$

Step 4: Execute the formula.

$$|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|$$

$$=70+28+46-14-23-9+4=144-46+4=102$$

Step 5: Take the complement. That 102 is the number who play at least one sport, so:

$$140-102=38$$

Watch the trap: 102 is sitting right there as option 1. It is the union, not the answer - the question asks for everyone *outside* it.

Final Answer: 38

The Setup: Pure function evaluation. No theory required - just substitute from the outside in, respecting each definition exactly as written.
Step 1: Set up the outer function. The definition is $g(y)=2y+f(y)$, so at $y=3$:

$$g(3)=2(3)+f(3)=6+f(3)$$

Note the variable name is a red herring: $f$ is written in $x$ and $g$ in $y$, but a function does not care what its argument is called. Feeding 3 into $f$ is the same operation either way.
Step 2: Evaluate the nested function. Using $f(x)=x^2+\log_3 x$ at $x=3$:

$$f(3)=3^2+\log_3 3$$

We have $3^2=9$, and any log of its own base is 1, since $\log_b b=1$ means *what power of $b$ gives $b$?* - namely 1. So $\log_3 3=1$ and $x=3>0$ keeps the log defined.

$$f(3)=9+1=10$$

Step 3: Recombine. Substitute back:

$$g(3)=6+10=16$$

Final Answer: 16

The Setup: A $2\times 2$ matrix $M=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is singular when its determinant vanishes, i.e. $ad-bc=0$, so we need the main-diagonal product to equal the anti-diagonal product: $ad=bc$. The whole question reduces to finding which four numbers from $\{1,2,3,4,5,6\}$ can be split into two disjoint pairs of equal product.
Step 1: Calculate the total sample space. We pick 4 distinct numbers from 6 and place them in 4 labelled positions, so order matters - this is a permutation:

$$\text{Total outcomes}=P(6,4)=6\times 5\times 4\times 3=360$$

Step 2: Find every pair of pairs with equal products. Rather than assert it, list all 15 pairs and their products outright:

Exactly two product values repeat, and in both cases the two pairs are disjoint (they share no element, which is required since all four entries must be distinct):
* Product $6$: $\{1,6\}$ and $\{2,3\}$
* Product $12$: $\{2,6\}$ and $\{3,4\}$

Every other product in the table is unique, so no further singular matrices exist.
Step 3: Count matrix arrangements. Take product 6. Put $\{1,6\}$ on the main diagonal $(a,d)$ and $\{2,3\}$ on the anti-diagonal $(b,c)$:
* Arrangements of $\{1,6\}$ across $a,d$: $2!=2$
* Arrangements of $\{2,3\}$ across $b,c$: $2!=2$

That gives $2\times 2=4$ matrices. But the two pairs can also switch diagonals - $\{2,3\}$ on the main, $\{1,6\}$ on the anti - for another 4. So product 6 contributes $8$ matrices, and by identical logic so does product 12.
Step 4: Final probability.

$$\text{Singular matrices}=8+8=16 \implies P=\frac{16}{360}=\frac{2}{45}$$

Final Answer: $\frac{2}{45}$

The Setup: In partnership accounting, profit is shared in proportion to each partner's *capital-months* - the money put in multiplied by how long it stayed in. Two partners with the same capital but different durations do not split evenly, and that is the whole mechanism here.
Step 1: Define Ashok's stats. Let Ashok's investment be $2x$ (choosing $2x$ rather than $x$ keeps Bharat's half a whole number). He started the business, so his money was in for the full 12 months:

$$\text{Ashok's capital-months}=2x\times 12=24x$$

Step 2: Define Bharat's stats. Bharat invested half of Ashok's amount, so his capital is $x$. Let it stay in for $t$ months:

$$\text{Bharat's capital-months}=x\times t=xt$$

Step 3: Equate to the profit ratio. Profit splits $3:1$ in the order the partners are named, Ashok to Bharat:

$$\frac{\text{Ashok's capital-months}}{\text{Bharat's capital-months}}=\frac{3}{1} \implies \frac{24x}{xt}=3$$

Step 4: Solve for the duration. Capital is positive, so $x$ cancels safely:

$$\frac{24}{t}=3 \implies 3t=24 \implies t=8$$

Step 5: Convert duration into a joining date - do not stop at Step 4. The number 8 is how long Bharat's money was *invested*, but the question asks when he joined. In a 12-month year, being invested for the last 8 months means he sat out the first four:

$$12-8=4\text{ months}$$

Answering 8 here is the intended trap, and it is why the question is phrased 'Bharat joined Ashok after' rather than asking for his investment period.

Final Answer: 4 months

The Setup: A min-maxing puzzle wearing a statistics costume. Sort the six marks as $S_1>S_2>S_3>S_4>S_5>S_6$, so the second highest is $S_2$ and the second lowest is $S_5$. We want to maximise $S_2-S_5$ subject to a fixed total, all marks distinct integers of at least 40, and one of them being exactly 70.
Step 1: Set the baseline constraints. Six students averaging 64 gives a fixed budget:

$$\text{Total}=6\times 64=384$$

Step 2: Pin the floor. $S_6\geq 40$, and since the marks are distinct integers, $S_5\geq S_6+1\geq 41$. So $S_5=41$ is the best we can ever do, achieved by putting $S_6=40$. Every point saved at the bottom is a point available at the top, so this choice never costs us anything.
Step 3: Decide where the mandatory 70 goes. This is the real work, and it needs all cases checked rather than a lucky guess:
* **70 as $S_1$ or $S_2$:** then $S_2\leq 70$, so the difference is at most $70-41=29$. Dead.
* **70 as $S_5$ or $S_6$:** impossible. If $S_5=70$ then $S_1,\dots,S_4$ all exceed 70, so they total at least $71+72+73+74=290$; adding $70$ and $S_6\geq 40$ already gives $400>384$. Placing 70 even lower is worse still.
* **70 as $S_4$:** then $S_3\geq 71$, and with $S_5=41,S_6=40$ the top three share $384-70-41-40=233$. Minimising $S_3$ at 71 leaves $S_1+S_2=162$, and $S_1>S_2$ forces $S_2\leq 80$. Difference at most $80-41=39$. Still beaten.
* **70 as $S_3$:** the surviving case, examined next.